Why nucleophilic substitution

Nu - could, for example, be OH - or CN -. In the process the electrons in the C-Br bond will be pushed even closer towards the bromine, making it increasingly negative.

Note: A co-ordinate bond is a covalent bond in which both electrons come from one of the atoms. The movement goes on until the -Nu is firmly attached to the carbon, and the bromine has been expelled as a Br - ion. Note: We haven't shown all the lone pairs on the bromine. These other lone pairs aren't involved in the reaction, and including them simply clutters the diagram to no purpose.

The large bromine atom hinders attack from its side and, being -, would repel the incoming Nu - anyway. This attack from the back is important if you need to understand why tertiary halogenoalkanes have a different mechanism. We'll discuss this later on this page. There is obviously a point in which the Nu - is half attached to the carbon, and the C-Br bond is half way to being broken. This is called a transition state. It isn't an intermediate. You can't isolate it - even for a very short time.

It's just the mid-point of a smooth attack by one group and the departure of another. Note: In exam you must show the lone pair of electrons on the nucleophile in this case, the Nu - ion. It probably doesn't matter whether you show them on the departing Br - ion or not. If you aren't happy about the use of curly arrows in mechanisms, follow this link before you go on.

Use the BACK button on your browser to return to this page. Technically, this is known as an S N 2 reaction. S stands for substitution, N for nucleophilic, and the 2 is because the initial stage of the reaction involves two species - the bromoethane and the Nu - ion. If your syllabus doesn't refer to S N 2 reactions by name, you can just call it nucleophilic substitution.

Some examiners like you to show the transition state in the mechanism, in which case you need to write it in a bit more detail - showing how everything is arranged in space. Be very careful when you draw the transition state to make a clear difference between the dotted lines showing the half-made and half-broken bonds, and those showing the bonds going back into the paper.

Notice that the molecule has been inverted during the reaction - rather like an umbrella being blown inside-out. Note: If you aren't happy about the various ways of drawing bonds , it is important to follow this link to find out exactly what the various symbols mean.

It is also important to know which of these ways of drawing the mechanism your particular examiners want you to use. If you haven't already checked your syllabus, recent exam papers and mark schemes , you must do so! At the time of writing, Edexcel, for example, wanted the transition state included, and that isn't obvious from their syllabus.

You have to check mark schemes and examiners reports. Check your syllabus, past papers and any support material published by your examiners to find out whether you need this. If there's no mention of tertiary halogenoalkanes or S N 1 reactions, then you probably don't need it.

Remember that a tertiary halogenoalkane has three alkyl groups attached to the carbon with the halogen on it. These alkyl groups can be the same or different, but in this section, we shall just consider a simple one, CH 3 3 CBr - 2-bromomethylpropane.

Once again, we'll talk this mechanism through using an ion as a nucleophile, because it's slightly easier, and again we'll look at the reaction of a general purpose nucleophilic ion which we'll call Nu -. With a tertiary halogenoalkane, this is impossible.

The back of the molecule is completely cluttered with CH 3 groups. We also need to consider the bulkiness of the substituents. Technically, this is known as an S N 2 reaction. S stands for substitution, N for nucleophilic, and the 2 is because the initial stage of the reaction involves two species - the bromoethane and the Nu - ion.

If your syllabus doesn't refer to S N 2 reactions by name, you can just call it nucleophilic substitution. Some examiners like you to show the transition state in the mechanism, in which case you need to write it in a bit more detail - showing how everything is arranged in space. Be very careful when you draw the transition state to make a clear difference between the dotted lines showing the half-made and half-broken bonds, and those showing the bonds going back into the paper.

Notice that the molecule has been inverted during the reaction - rather like an umbrella being blown inside-out. Remember that a tertiary halogenoalkane has three alkyl groups attached to the carbon with the halogen on it. These alkyl groups can be the same or different, but in this section, we shall just consider a simple one, CH 3 3 CBr - 2-bromomethylpropane. Once again, we'll talk this mechanism through using an ion as a nucleophile, because it's slightly easier, and again we'll look at the reaction of a general purpose nucleophilic ion which we'll call Nu -.

With a tertiary halogenoalkane, this is impossible. The back of the molecule is completely cluttered with CH 3 groups. Since any other approach is prevented by the bromine atom, the reaction has to go by an alternative mechanism.

The reaction happens in two stages. In the first, a small proportion of the halogenoalkane ionises to give a carbocation and a bromide ion. This reaction is possible because tertiary carbocations are relatively stable compared with secondary or primary ones. Even so, the reaction is slow. Once the carbocation is formed, however, it would react immediately it came into contact with a nucleophile like Nu -. The lone pair on the nucleophile is strongly attracted towards the positive carbon, and moves towards it to create a new bond.

How fast the reaction happens is going to be governed by how fast the halogenoalkane ionises. Because this initial slow step only involves one species, the mechanism is described as S N 1 - substitution, nucleophilic, one species taking part in the initial slow step.

If a primary halogenoalkane did use this mechanism, the first step would be, for example:. A primary carbocation would be formed, and this is much more energetically unstable than the tertiary one formed from tertiary halogenoalkanes - and therefore much more difficult to produce. This instability means that there will be a very high activation energy for the reaction involving a primary halogenoalkane. The activation energy is much less if it undergoes an S N 2 reaction - and so that's what it does instead.

There isn't anything new in this. Secondary halogenoalkanes will use both mechanisms - some molecules will react using the S N 2 mechanism and others the S N 1. The S N 1 mechanism is possible because the secondary carbocation formed in the slow step is more stable than a primary one.

It isn't as stable as a tertiary one though, and so the S N 1 route isn't as effective as it is with tertiary halogenoalkanes. Jim Clark Chemguide. Jonathan Mooney McGill University. The strengths of the carbon-halogen bonds In all of these nucleophilic substitution reactions, the carbon-halogen bond has to be broken at some point during the reaction. Nucleophilic substitution in primary halogenoalkanes You will need to know about this if your syllabus talks about "primary halogenoalkanes" or about S N 2 reactions.

The nucleophilic substitution reaction - an S N 2 reaction We'll talk this mechanism through using an ion as a nucleophile, because it's slightly easier. Nucleophile effects A major factor affecting the S N 2 reaction is strength of the nucleophile. We can describe several general trends which determine the strength of a nucleophile: A molecule which has lost a proton and is negatively charged base is stronger than the neutral, protonated version of the same molecule conjugate acid.

Nucleophilicity increases in parallel with the base strength. Thus, amines, alcohols and alkoxides are very good nucleophiles. Base strength is a rough measure of how reactive the nonbonding electron pair is; thus, it is not necessary for a nucleophile to be anionic. Under substitution conditions, amines proceed all the way to form quaternary salts, which makes it difficult to control the extent of the reaction. However, as a nucleophile's base strength and steric hindrance increase, its basicity tends to be accentuated.

An additional factor that plays a role is the character of the solvent. Increasing stabilization of the nucleophile by the solvent results in decreasing reactivity. Thus, polar protic solvents will stabilize the chloride and bromide ions through the formation of hydrogen bonds to these smaller anions. Iodide is a comparatively better nucleophile in these solvents. The reverse behavior predominates in aprotic polar media.

The solvent also plays an important role in determining which pathway the reaction will take, S N 1 versus S N 2. It may safely be assumed that a primary-substituted leaving group will follow an S N 2 pathway in any case, since the formation of the corresponding unstable primary carbenium ion is disfavored. Reaction by the S N 1 pathway is highly probable for compounds with tertiary substitution, since the corresponding tertiary carbenium ion is stabilized through hyperconjugation:.

The better the solvent stabilizes the ions, the more probable that the reaction will follow an S N 1 pathway e. The more highly substituted is the incipient carbenium ion, the more probable that the reaction will follow an S N 1 pathway. The more unreactive the nucleophile, the more probable it becomes that a reaction with secondary and tertiary electrophiles will follow an S N 1 pathway.

A weaker nucleophile is not as effective in the backside attack, since this location is sterically shielded, especially in the case of tertiary substrates. Carbenium ions are planar and therefore less sterically hindered, and are naturally more reactive as electrophiles than the uncharged parent compound. Kim, C. Song, D. Chi, J. Jadhav, J. Kim, H.